## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 235: 40

#### Answer

The simplified answer of the given expression is ${{1010010}_{\text{two}}}$.

#### Work Step by Step

Since, the given computation involves base two, the only allowed numbers are 0 and 1. Break the given computation into two parts. First add ${{11100}_{\text{two}}}+{{11111}_{\text{two}}}$ and then add the result obtained to the third number in the given expression, ${{10111}_{\text{two}}}$. To add two numbers, first write the two numbers one over the other, such that the digits corresponding to the same place value come in one line and then add column by column: ${{11100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ Now, add the two digits in the first column from right: ${{0}_{\text{two}}}+{{1}_{\text{two}}}={{1}_{\text{ten}}}=\left( 1\times 1 \right)={{1}_{\text{two}}}$. Write ${{1}_{\text{two}}}$ in the first column from right below the horizontal line: ${{11100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ ${{1}_{\text{two}}}$ Now, add the two digits in the second column from right: ${{0}_{\text{two}}}+{{1}_{\text{two}}}={{1}_{\text{ten}}}=\left( 1\times 1 \right)={{1}_{\text{two}}}$. Write ${{1}_{\text{two}}}$ in the second column from right below the horizontal line: ${{11100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ ${{11}_{\text{two}}}$ Now, add the two digits in the third column from right: ${{1}_{\text{two}}}+{{1}_{\text{two}}}={{2}_{\text{ten}}}=\left( 1\times 2 \right)+\left( 0\times 1 \right)={{10}_{\text{two}}}$. Write ${{0}_{\text{two}}}$ in the third column from right below the horizontal line, and carry ${{1}_{\text{two}}}$ to the fourth column from right: $1\overset{1}{\mathop{1}}\,{{100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ ${{011}_{\text{two}}}$ Now, add the three digits in the fourth column from right: ${{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}={{3}_{\text{ten}}}=\left( 1\times 2 \right)+\left( 1\times 1 \right)={{11}_{\text{two}}}$. Write ${{1}_{\text{two}}}$ in the fourth column from right below the horizontal line, and carry ${{1}_{\text{two}}}$ to the fifth column from right: $\overset{1}{\mathop{1}}\,{{1100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ ${{1011}_{\text{two}}}$ Now, add the three digits in the fifth column from right: ${{1}_{\text{two}}}+{{1}_{\text{two}}}+{{1}_{\text{two}}}={{3}_{\text{ten}}}=\left( 1\times 2 \right)+\left( 1\times 1 \right)={{11}_{\text{two}}}$. Write ${{11}_{\text{two}}}$ in front of ${{1011}_{\text{two}}}$ below the horizontal line: ${{11100}_{\text{two}}}$ $\underline{+{{11111}_{\text{two}}}}$ ${{111011}_{\text{two}}}$ Now, add the result obtained above to ${{10111}_{\text{two}}}$: ${{111011}_{\text{two}}}$ $\underline{+{{10111}_{\text{two}}}}$ Applying the same procedure used in the addition ${{11100}_{\text{two}}}+{{11111}_{\text{two}}}$, the above addition can be computed. The result will be: ${{111011}_{\text{two}}}$ $\underline{+{{10111}_{\text{two}}}}$ ${{1010010}_{\text{two}}}$ Now, to check whether the above obtained answer is correct, compute the given expression by converting each number to base ten: ${{11100}_{\text{two}}}=28$, ${{11111}_{\text{two}}}=31$, ${{10111}_{\text{two}}}=23$ and ${{1010010}_{\text{two}}}=82$. Since, $28+31+23$ indeed equals 82, the solution obtained is correct. Hence, the simplified answer of the given expression is ${{1010010}_{\text{two}}}$.

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